1 /*************************************************************************
  2     > File Name: hdu-2389.rain_on_your_parade.cpp
  3     > Author: CruelKing
  4     > Mail: 2016586625@qq.com 
  5     > Created Time: 2019年09月02日 星期一 21时29分16秒
  6     本题思路:比较裸的二分匹配,但是一看n比较大,所以需要更牛皮的算法,也即HK算法其复杂度为sqrt(n) * m.
  7  ************************************************************************/
  8 
  9 #include <cstdio>
 10 #include <cmath>
 11 #include <cstring>
 12 #include <vector>
 13 #include <queue>
 14 #include <map>
 15 using namespace std;
 16 
 17 typedef long long ll;
 18 const int maxn = 3000 + 5, inf = 0x3f3f3f3f;
 19 int speed[maxn];
 20 bool used[maxn];
 21 int n, m, t;
 22 typedef pair<int, int> pii;
 23 pii umbrellas[maxn], guests[maxn];
 24 int mx[maxn], my[maxn];
 25 int dx[maxn], dy[maxn];
 26 vector <int> G[maxn];
 27 int dis;
 28 
 29 bool searchp() {
 30     queue<int> que;
 31     dis = inf;
 32     memset(dx, -1, sizeof dx);
 33     memset(dy, -1, sizeof dy);
 34     for(int i = 1; i <= m; i ++) {
 35         if(mx[i] == -1) {
 36             que.push(i);
 37             dx[i] = 0;
 38         }
 39     }
 40     while(!que.empty()) {
 41         int u = que.front();
 42         que.pop();
 43         if(dx[u] > dis) break;
 44         int sz = G[u].size();
 45         for(int i = 0; i < sz; i ++) {
 46             int v = G[u][i];
 47             if(dy[v] == -1) {
 48                 dy[v] = dx[u] + 1;
 49                 if(my[v] == -1) dis = dy[v];
 50                 else {
 51                     dx[my[v]] = dy[v] + 1;
 52                     que.push(my[v]);
 53                 }
 54             }
 55         }
 56     }
 57     return dis != inf;
 58 }
 59 
 60 bool dfs(int u) {
 61     int sz = G[u].size();
 62     for(int i = 0; i < sz;i  ++) {
 63         int v = G[u][i];
 64         if(!used[v] && dy[v] == dx[u] + 1) {
 65             used[v] = true;
 66             if(my[v] != -1 && dy[v] == dis) continue;
 67             if(my[v] == -1 || dfs(my[v])) {
 68                 my[v] = u;
 69                 mx[u] = v;
 70                 return true;
 71             }
 72         }
 73     }
 74     return false;
 75 }
 76 
 77 int maxmatch() {
 78     int res = 0;
 79     memset(mx, -1, sizeof mx);
 80     memset(my, -1, sizeof my);
 81     while(searchp()) {
 82         memset(used, false, sizeof used);
 83         for(int i = 1; i <= m;i ++) {
 84             if(mx[i] == -1 && dfs(i)) res ++;
 85         }
 86     }
 87     return res;
 88 }
 89 
 90 bool has_distance(int i, int j) {
 91     ll temp = (umbrellas[i].first - guests[j].first) * (umbrellas[i].first - guests[j].first) + (umbrellas[i].second - guests[j].second) * (umbrellas[i].second - guests[j].second);
 92     return temp <= (ll)speed[j] * speed[j] * t * t;
 93 }
 94 
 95 
 96 
 97 int main() {
 98     int T, Case = 0;
 99     scanf("%d", &T);
100     while(T --) {
101         scanf("%d", &t);
102         scanf("%d", &m);
103         for(int i = 1; i <= m; i ++) {
104             scanf("%d %d %d", &guests[i].first, &guests[i].second, &speed[i]);
105         }
106         scanf("%d", &n);
107         for(int i = 1; i <= n; i ++) {
108             scanf("%d %d", &umbrellas[i].first, &umbrellas[i].second);
109             for(int j = 1; j <= m; j ++) {
110                 if(has_distance(i, j)) G[j].push_back(i);
111             }
112         }
113         int res = maxmatch();
114         for(int i = 1; i <= m; i ++) G[i].clear();
115         printf("Scenario #%d:\n", ++Case);
116         printf("%d\n\n", res);
117     }
118     return 0;
119 }